3.1.29 \(\int \frac {3+x+5 x^2+(30 x+10 x^2) \log (6+2 x)}{3 x+x^2+(15 x^2+5 x^3) \log (6+2 x)} \, dx\)

Optimal. Leaf size=16 \[ \log \left (-3 x-15 x^2 \log (2 (3+x))\right ) \]

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Rubi [A]  time = 0.24, antiderivative size = 14, normalized size of antiderivative = 0.88, number of steps used = 2, number of rules used = 2, integrand size = 52, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.038, Rules used = {6741, 6685} \begin {gather*} \log (x (5 x \log (2 (x+3))+1)) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(3 + x + 5*x^2 + (30*x + 10*x^2)*Log[6 + 2*x])/(3*x + x^2 + (15*x^2 + 5*x^3)*Log[6 + 2*x]),x]

[Out]

Log[x*(1 + 5*x*Log[2*(3 + x)])]

Rule 6685

Int[(u_)/((w_)*(y_)), x_Symbol] :> With[{q = DerivativeDivides[y*w, u, x]}, Simp[q*Log[RemoveContent[y*w, x]],
x] /;  !FalseQ[q]]

Rule 6741

Int[u_, x_Symbol] :> With[{v = NormalizeIntegrand[u, x]}, Int[v, x] /; v =!= u]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {3+x+5 x^2+\left (30 x+10 x^2\right ) \log (6+2 x)}{x (3+x) (1+5 x \log (2 (3+x)))} \, dx\\ &=\log (x (1+5 x \log (2 (3+x))))\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.21, size = 15, normalized size = 0.94 \begin {gather*} \log (x)+\log (1+5 x \log (2 (3+x))) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(3 + x + 5*x^2 + (30*x + 10*x^2)*Log[6 + 2*x])/(3*x + x^2 + (15*x^2 + 5*x^3)*Log[6 + 2*x]),x]

[Out]

Log[x] + Log[1 + 5*x*Log[2*(3 + x)]]

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fricas [A]  time = 0.70, size = 21, normalized size = 1.31 \begin {gather*} 2 \, \log \relax (x) + \log \left (\frac {5 \, x \log \left (2 \, x + 6\right ) + 1}{x}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((10*x^2+30*x)*log(2*x+6)+5*x^2+x+3)/((5*x^3+15*x^2)*log(2*x+6)+x^2+3*x),x, algorithm="fricas")

[Out]

2*log(x) + log((5*x*log(2*x + 6) + 1)/x)

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giac [A]  time = 0.37, size = 15, normalized size = 0.94 \begin {gather*} \log \left (5 \, x \log \left (2 \, x + 6\right ) + 1\right ) + \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((10*x^2+30*x)*log(2*x+6)+5*x^2+x+3)/((5*x^3+15*x^2)*log(2*x+6)+x^2+3*x),x, algorithm="giac")

[Out]

log(5*x*log(2*x + 6) + 1) + log(x)

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maple [A]  time = 0.05, size = 16, normalized size = 1.00

method	result	size
norman	\(\ln \relax (x )+\ln \left (5 x \ln \left (2 x +6\right )+1\right )\)	\(16\)
risch	\(2 \ln \relax (x )+\ln \left (\ln \left (2 x +6\right )+\frac {1}{5 x}\right )\)	\(19\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((10*x^2+30*x)*ln(2*x+6)+5*x^2+x+3)/((5*x^3+15*x^2)*ln(2*x+6)+x^2+3*x),x,method=_RETURNVERBOSE)

[Out]

ln(x)+ln(5*x*ln(2*x+6)+1)

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maxima [A]  time = 0.76, size = 25, normalized size = 1.56 \begin {gather*} 2 \, \log \relax (x) + \log \left (\frac {5 \, x \log \relax (2) + 5 \, x \log \left (x + 3\right ) + 1}{5 \, x}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((10*x^2+30*x)*log(2*x+6)+5*x^2+x+3)/((5*x^3+15*x^2)*log(2*x+6)+x^2+3*x),x, algorithm="maxima")

[Out]

2*log(x) + log(1/5*(5*x*log(2) + 5*x*log(x + 3) + 1)/x)

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mupad [B]  time = 0.40, size = 15, normalized size = 0.94 \begin {gather*} \ln \left (5\,x\,\ln \left (2\,x+6\right )+1\right )+\ln \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x + log(2*x + 6)*(30*x + 10*x^2) + 5*x^2 + 3)/(3*x + log(2*x + 6)*(15*x^2 + 5*x^3) + x^2),x)

[Out]

log(5*x*log(2*x + 6) + 1) + log(x)

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sympy [A]  time = 0.21, size = 17, normalized size = 1.06 \begin {gather*} 2 \log {\relax (x )} + \log {\left (\log {\left (2 x + 6 \right )} + \frac {1}{5 x} \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((10*x**2+30*x)*ln(2*x+6)+5*x**2+x+3)/((5*x**3+15*x**2)*ln(2*x+6)+x**2+3*x),x)

[Out]

2*log(x) + log(log(2*x + 6) + 1/(5*x))

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